Question

我有两个数据集， partb_data1 和 partb_data2 。给出反映客户特征的银行客户样本以及银行是否继续与他们合作（流失）。 退出：Churn（如果他离开了银行，则为 1，如果他继续与银行合作，则为 0）。 我使用 partb_data1 作为训练集，partb_data2 作为测试集。

这是我的数据：

> dput(head(partb_data1))
structure(list(RowNumber = 1:6, CustomerId = c(15634602L, 15647311L, 
15619304L, 15701354L, 15737888L, 15574012L), Surname = c("Hargrave", 
"Hill", "Onio", "Boni", "Mitchell", "Chu"), CreditScore = c(619L, 
608L, 502L, 699L, 850L, 645L), Geography = c("France", "Spain", 
"France", "France", "Spain", "Spain"), Gender = c("Female", "Female", 
"Female", "Female", "Female", "Male"), Age = c(42L, 41L, 42L, 
39L, 43L, 44L), Tenure = c(2L, 1L, 8L, 1L, 2L, 8L), Balance = c(0, 
83807.86, 159660.8, 0, 125510.82, 113755.78), NumOfProducts = c(1L, 
1L, 3L, 2L, 1L, 2L), HasCrCard = c(1L, 0L, 1L, 0L, 1L, 1L), IsActiveMember = c(1L, 
1L, 0L, 0L, 1L, 0L), EstimatedSalary = c(101348.88, 112542.58, 
113931.57, 93826.63, 79084.1, 149756.71), Exited = c(1L, 0L, 
1L, 0L, 0L, 1L)), row.names = c(NA, 6L), class = "data.frame")



> dput(head(partb_data2))
structure(list(RowNumber = 8001:8006, CustomerId = c(15629002L, 
15798053L, 15753895L, 15595426L, 15645815L, 15632848L), Surname = c("Hamilton", 
"Nnachetam", "Blue", "Madukwe", "Mills", "Ferrari"), CreditScore = c(747L, 
707L, 590L, 603L, 615L, 634L), Geography = c("Germany", "Spain", 
"Spain", "Spain", "France", "France"), Gender = c("Male", "Male", 
"Male", "Male", "Male", "Female"), Age = c(36L, 32L, 37L, 57L, 
45L, 36L), Tenure = c(8L, 9L, 1L, 6L, 5L, 1L), Balance = c(102603.3, 
0, 0, 105000.85, 0, 69518.95), NumOfProducts = c(2L, 2L, 2L, 
2L, 2L, 1L), HasCrCard = c(1L, 1L, 0L, 1L, 1L, 1L), IsActiveMember = c(1L, 
0L, 0L, 1L, 1L, 0L), EstimatedSalary = c(180693.61, 126475.79, 
133535.99, 87412.24, 164886.64, 116238.39), Exited = c(0L, 0L, 
0L, 1L, 0L, 0L)), row.names = c(NA, 6L), class = "data.frame")

我创建了分类树以预测流失。代码如下：

library(tidyverse)
library(caret)
library(rpart)
library(rpart.plot)

# Split the data into training and test set
train.data <- head(partb_data1, 500)
test.data <- tail(partb_data2, 150)

# Build the model
modelb <- rpart(Exited ~., data = train.data, method = "class")
# Visualize the decision tree with rpart.plot
rpart.plot(modelb)

# Make predictions on the test data
predicted.classes <- modelb %>% 
  predict(test.data, type = "class")
head(predicted.classes)

# Compute model accuracy rate on test data
mean(predicted.classes == test.data$Exited)

### Pruning the tree :

# Fit the model on the training set
modelb2 <- train(
  Exited ~., data = train.data, method = "rpart",
  trControl = trainControl("cv", number = 10),
  tuneLength = 10
)
# Plot model accuracy vs different values of
# cp (complexity parameter)
plot(modelb2)

# Print the best tuning parameter cp that
# maximizes the model accuracy
modelb2$bestTune

# Plot the final tree model
plot(modelb2$finalModel)

# Make predictions on the test data
predicted.classes <- modelb2 %>% predict(test.data)
# Compute model accuracy rate on test data
mean(predicted.classes == test.data$Exited)

注意：我已经从 partb_data2 制作了测试集。

我遵循的程序正确吗？我必须进行任何更改才能实现我的目标，即分类树？非常欢迎您的帮助！

已编辑！！！

Answer 1

你的 head(partb_data1$Exited, 500) 不是 data.frame。由于 $，您需要获取 partb_data1 数据的子集。它只是一个整数向量，所以行不通。

<块引用>

class(head(partb_data1$Exited, 500)) [1] "整数"

Answer 2

总是有很多程序选项。

但是您将数据分离为训练数据集和测试数据集是正确的。也可以改用交叉验证。您正在训练集上使用交叉验证，这通常不是必需的，但也是可能的。

我认为将您的完整数据用于 cv 也应该有效，但您所做的并没有错。

构建决策树分类

2 个答案: